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How many parameters are stored in the flight data recorder these days?
Originally, it was seven parameters, scratched into foil tape in analog form.
Many years later, a video said it was over 100 parameters stored digitally. About 10 years ago, someone referred to it being several hundred.
How many do they store now? Do they store literally every piece of information available, including trivial things like whether the lavatory door is locked?
Is there a master list I could look at? I can't imagine it's secret.

I do not think there is a master list (aka unique list). Now it is "mostly" an hard disk: Aircraft manufacturers may include what they may think interesting. FAA have minimal requirements (depending on aircraft type) (and NTSB has additional recommendations, which I assume somebody will add as answer).

For aircraft following FAA regulations, modern FDRs for air carriers must store 88 parameters, which are listed in 14 CFR 135 Appendix D. The previous lists are in 135.152, there were 11 parameters for pre-1969 aircraft and 29 for post-1969. The list was expanded in 1991.
In reality "over 1000" parameters are actually recorded, NTSB:

By regulation, newly manufactured aircraft must monitor at least eighty-eight important parameters such as time, altitude, airspeed, heading, and aircraft attitude. In addition, some FDRs can record the status of more than 1,000 other in-flight characteristics that can aid in the investigation.

Other smaller lists can be seen in 125.225 125.226, they contain 11, 17, and 91 entries. 125 Appendix D has another list. These specify the FDR regulations for air planes, vs air carriers in Part 135.
And finally, AC 20-141B discusses the technical details of a digital FDR. There are some appendices that may be helpful, such as determining the requirements for various categories of air carriers and airplanes.
Some similar previous questions:

Is there anywhere I can pick up raw flight data recorder information?
What is the format of data stored by a Flight Data Recorder?
How much data is stored in the flight data recorder per hour in a typical flight of an A380?

GOD! I wish all answers here were as complete and great! I followed your links to discover the Fred format that's used on flight data recorders. https://papers.phmsociety.org/index.php/phmconf/article/download/1400/373 ===== THANK YOU!

It should be noted that these parameters are logged at different rates. While the basic flight parameters are logged with the highest rate possible, other less important parameters might only be logged once a second or even once a minute (e. G. Door status) in order to save data storage space


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Instead of battery why we don't use some perpetual energy source? Atomic Energy?
Right now the biggest issue of the electric plane is carrying a large battery to store energy proportional to its consumption for long-range, to increase range we use a larger battery. Even solar planes need batteries.
Why not develop some kind of small reactor that generates energy and has a very compact size. Reducing the size of battery energy direct transfer to the motors.

While I understand the context of your question here, I think this one is better for Engineering or Physics SEs.  The short answer is "regulatory/political forces prevent nuclear reactors in private hands" but the longer answer is that you're basically asking "Why don't we just invent a better mousetrap?"

Nuclear power is ultimately steam power. Guess why nobody has tried building a steam airplane?

You build an actual, functioning perpetual energy source and the world will beat a path to your door step.

Because there is no such thing as a perpetual energy source. Atomic energy eventually runs out and all reactors have to be refueled. Stars all run out of fuel as well and flame out in some interesting, and often spectacular way. Putting a nuclear reactor in an aircraft has already been explored and it was found that properly shielding it would make the airframe too heavy to fly.
Why nuclear powered aircraft do not exist.

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When told "call sign only" in a frequency change, how to transmit it?
For example: London Control says:

BAW211 contact London Approach 118.1, call sign only.

What will the pilot say to approach on 118.1?

London Approach, BAW211.

Or just:

BAW211.

What info are they asking you to leave off?  In my experience, the "full" call would've only been "London Approach, BAW211, 4500, descending".  I cannot believe that adding altitude and v-speed contributes to "channel congestion".

@abelenky According to the UK's Radiotelephony Manual, an example for an initial call would be "Westbury Departure, BIGJET 347, BIGRO 5D, Passing Altitude 2300 feet climbing FL80", so more than twice as long as callsign only.

... And having experienced the LHR approach a number of times, it gets very busy there...

It's used outside of LHR, and it's not even when things are that busy. I get handed over to Southend Director with a 'callsign only' when it's dead.

@GdD if they’ve got systems in place to handle it when it’s busy, there’s not much reason to turn those systems off.

You always start a communication with the intended receiver, so in this case it would be London Approach, BAW211. What the controller is telling BAW211 is that London Approach has the relevant details of the flight so BAW doesn't need to communicate anything further or make a request of london approach. This helps keep the frequency clear of redundant communications.
As for why you always lead with the recipient in the communication, not only is it in the regs it just makes sense. Think of it this way, if you get on the radio and say 'It's Frank.' nobody will know who you're taking to, if you say 'Mary, it's Frank.' then it's clear who you are speaking to.

also if you open with BAW211 other listeners will think the message is for you, how will approach know it's addressed to them?

A very good point @Jasen.

You always include the identifier of the addressee, so London Approach, Speedbird 211.
This triggers the air traffic control officer, who may be mentally busy with something else, to listen. If the London Approach bit is left out, there is an increased chance of the whole message being missed or misheard, leading to a need for additional communication which the call sign only method was designed to prevent.

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How does an aircraft work with ILS
In what situation will an aircraft show you "LOC*" in FMA?
What signal does the aircraft received when LOC*?(how does the aircraft know it is on LOC*)
And which heading will the aircraft choose,is there any formula for the heading?

The aircraft is a machine, incapable of making a choice.  Generally the pilot will choose the heading to fly.  The formula is zero plus the number of degrees of the chosen heading.

You're asking multiple questions at once, some of them quite broad. I suggest you do some reading on Instrument Landing System or the Localizer and then come back to ask specific questions when something is still unclear.

The Localizer signal is the horizontal (lateral) component of the full ILS signal (top portion of the image). It consists of two "lobes" which are amplitude modulated from each other, allowing the receiver to distinguish between left and right. The FMA reporting LOC has identified what it believes to be the ILS localizer signal, but NOT the glideslope signal (which is the same technique, but rotated 90-degrees so that the system is able to distinguish up from down, instead). You have side-to-side guidance along the runway centerline, but no guidance as to what altitude you should be flying at.
LOC* (as opposed to LOC) specifically refers to the "capture" phase. It's mechanically no different from LOC mode, but having a specific mode for the capture phase allows for a different set of logic to control the capture vs. the hold phase of the approach.  The capture phase essentially exists between the moment you pick up the ILS signal until you've stabilized on the centerline the first time. After that I would expect the FMA to say "LOC" instead of "LOC*".
As to how the aircraft will maneuver itself in response to that signal, that will vary by aircraft model, possibly operator, and possibly even specific configurations of a given aircraft. Generally the heading for the capture maneuver is whatever vector you're given by Approach, until you've intercepted the beam itself. At that point the aircraft is turning to  match the approach heading for the given runway but will shift as needed to stay centered on the beam.  The maths for that are fairly complex.  But if you're curious, here's an example method using H-infinity synthesis, if you want to dig into it.

"LOC"=in the centre of the two "lobes".   "LOC*"=in one lobe,and at the   edge   of the other one,is that correct?   is it the same logic for glide path?

"LOC" = have been in the centre of the two lobes at some point; "LOC*" = consistently in one lobe and haven't yet seen the centre of both.
I would expect it's similar for glideslope, but the consequences of missing the GS too much downwards are MUCH more likely to be severe than right or left so I'd wager the programming is much more conservative in how it executes.


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Weight & Balance with multiple stops
When you have multiple stops for a single flight, without refuelling, how do you do deal with the weight and balance ?
For every single leg, or just from departure to arrival ?

Any other load changes at the stops? Passengers? Baggage? Cargo?

@757toga it's a GA aircraft, Cessna 172.

@FredLarson no changes, just a stop, to see the airport. And then the flight continues.

When you say 'deal' what are you asking?

Can you edit your question to add in those relevant clarifications?

From a private flying perspective there's no regulation stating you absolutely must do a weight and balance calculation before a flight, you have to ensure you are operating within the airplane's limitations.
I fly 172s and PA-28s mostly, usually just myself with just a flight bag, and I don't do a weight and balance calculation before every flight because I've done them for each airplane I fly already, and I know it will be within the envelope as nothing has changed. If I change those parameters, i.e. take more people or carry bags/cargo in the back I will do weight and balance. When I do the calculations I'll look at variable fuel loads so I will know how it will change as I burn fuel and ensure I'll be within limits for the duration of the flight, or be within limits for the return flight. So I won't do a weight and balance for the return leg unless something changes, for instance take on a passenger.
Performance calculations are not to be ignored, especially if you are operating close to weight limits, out of a short field, or at a high density altitude. You don't need to do the balance for that, just calculate the weight.

It's pretty simple on that type of aircraft if you're not adding or removing load at each stop. On the chart you simply calculate your CG without fuel (zero fuel weight), and with your max fuel load of the day. Draw a line between the two and if the whole line is in the envelope you know that each flight that day will be within limits. Carry that with you in case you get a ramp check so you can show you've considered the weight and balance for that flight.

Depends. The question is one of takeoff performance mostly.  I would do a W&B on the initial departure leg only, if my only weight changes on each leg are from fuel burn, and the subsequent legs are from airports with ample takeoff distance margin.
However, if an airport on my route is marginal from a takeoff distance perspective, my All Up Weight when taking off from that airport becomes critical and I would do a W&B adjustment to account for the reduced All Up Weight in determining if can get out of the strip in the first place.  But I would be doing that estimation before I departed on the first leg anyway, and probably revisit it when at the airport in question to make sure my assumptions still applied.
Otherwise, if you just hopping around from 4000 ft airport to 4000 ft airport, there's really no need.  In fact as a general practice I only do detailed walkarounds and engine runups on the first flight of the day, with more of a cursory once-over on subsequent legs to make sure nothing happened to it while it was parked.


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Why is the true proportional guidance law considered closed form? What does 'closed form' mean over here and how does it apply to other guidance laws?

Can someone please explain why they refer to the solution as closed-form? From my understanding, the term closed form would imply that the command constant 'c' can only take values from a finite set? Why so?
And how does it defer from Pure proportional navigation, pursuit navigation and LOS navigation in terms of how the solution is closed from or not?

In mathematical terms, an equation is of "closed form" if it can be solved exactly without simplifying assumptions, successive approximations or by for example discarding higher-order terms in a polynomial expansion of one of the state functions. Hence, an equation which admits a closed-form solution can be solved by an explicit algorithm containing a finite number of steps that could be programmed into the feedback block of an automatic control system, for example.


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How would you enter this approach with a racetrack course reversal at EGPH from the southwest?
For reference this is the NDB ILS DME RWY 24 approach at EGPH. When arriving from the south/southwest, I am sure you would be vectored in but i am interested as to how this would be done without vectors. Assuming you are not coming from TLA, this would mean you cannot use the DME arc so your inital approach fix would be EDN, requiring a course reversal on the racetrack.
A few points:

What is the smaller racetrack inside the larger one?
I am assuming this would require a parallel entry? If so, how long would you travel along the inbound section before turning back around?
Could you descend and intercept the glideslope after completing the racetrack/hold entry or would you have to do the entry, get back to EDN, then go back around the main racetrack to the top of the glideslope?

A hold is a hold. A racetrack is a racetrack.

Hmm. Pressed Enter too early! And I don't know how to format it...
A hold is a hold (but it does have a racetrack 'shape'!). A racetrack is a racetrack.
Note Racetracks are in bold. Holds are in normal (non-bold).
A hold is used as a delay.
A Racetrack is used to put you in a position where you can join the final approach course.

Hahaha I thought you were being snarky for a minute there! Thanks for the clarification

A caveat in that I haven't experienced the world of EASA, but this is how I see it through international convention:

What is the smaller racetrack inside the larger one?

This is just the normal holding pattern, albeit with left hand turns.

I am assuming this would require a parallel entry? If so, how long would you travel along the inbound section before turning back around?

Yes, and because it isn't otherwise specified, I would fly 1 minute legs.

Could you descend and intercept the glideslope after completing the racetrack/hold entry or would you have to do the entry, get back to EDN, then go back around the main racetrack to the top of the glideslope?

No. The minimum altitude in the hold is 3000ft, and after 1 minute the glideslope would still be well beneath you.
When you are ready to commence the approach, you simply exit the hold via an extension of the outbound leg, on a track of 061. You must maintain this track and altitude until 12DME. From there you can turn left and intercept the localiser, remaining at 3000 until you have the glideslope (which can be expected at 9DME, which is also the FAF).

Thank you, that confirms my suspicions! Follow up: Would an FMS be capable of flying this automatically or would this need to be dialled in manually?

The IAF is EDN, So, from the southwest I would approach the NDB, cross it, do a teardrop entry to the hold, cross the beacon again and start my procedural left turn.
When complete I would be on an outbound track of 061 degrees at which point I would report beacon outbound (if that was the instruction given - it iusually is). From there the procedure is fairly straightforward. Proceed out to 12DME and start left procedural turn to final track. Descend from 9DME with the glideslope.

There are not 2 racetracks. There is one standard hold which just so happens to be completely aligned with the outbound and inbound track.

I dont think this does require parallel entry, from the southwest I think you would be absolutely fine to do a teardrop entry as described above.

I dont quite follow your third question - but lets just say, in the real world, just ask for vectors to ILS :)

If you have a specific STAR that approaches you from south/south west and you are eager to apply this procedure, look at the plan view, chart does not say it is mandatory to approach from TLA side but it starts at over lctr at 3000'.
You follow your southern arrival procedure (star) descent 3000' and go over lctr, make parallel entry for the holding path and wait for your turn to approach. Get the 'clear ILS NDB aprooach for rwy 24'.Do approach checklist before you leave the holding. Fly outbound for 12nm turn for inbound do not descent until 9nm at which you will catch glide slope. Clear to land rwy 24!!

You are right that would be parallel !


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Could an executive configured ATR 42-600 get 2000nm range?
Say you wanted to outfit an ATR 42-600 for executive transportation like a smaller version of an Airbus ACJ 220. For example, 2 living areas of 4 rows, 1 sleeping area of 3 rows (a double below and a single above on both sides), and increase galley and lavatory by 1 row.
In commercial layout the maxpax range is 720nm. Could the the above executive config get close to 2000nm range? Looking at the King Air 360er range of 2500nm for such a small plane I'm thinking it could be possible. Less weight. Maybe convert some baggage area to fuel tanks. Possible?
Also, I know there is at least one VIP variant out there but it only converts about half of the cabin and leaves the other with economy seating. I imagine that would be heavier than the above.

Worth noting: executive configuration is not light... Plus th ATR is not one of the quietest airplanes, so something would need to be done to that too. And it's slow, so...

@Jpe61Not true on the noise. Modern turboprop cabins are just as quite as jets, both private and commercial.

Why the ATR?  Why not start with something that’s longer range to begin with?

@MichaelHall Cost and size. An ATR 42-600 is less than 10mil. So if it could be converted and get the 2000nm range for a couple of mil then you basically have a jumbo King Air for less than twice the cost.

@Edmund the ATR's I've travelled in absolutely cannot compete with , say, 737MAX, or don't come even close to A350... Noise, vibrations, no chance.

Well, between size, cost, and range, I think you will need to choose which is most important and accept a compromise...

No, it wouldn't be possible as the airplane cannot carry that much weight. The ATR 42-600 has a maximum take-off weight of 18,600kg, and an empty weight of 11,550kg, so its maximum load is about 7050kg. It's maximum fuel capacity is 4500kg which gives a range of 720nm, which equates to 6.25nm per kg of fuel. If we use that figure for a range of 2000nm the weight of the fuel is 12,500kg, which is 5000kg over the maximum weight. That 12,500 weight is an over-simplification as airplanes burn much more fuel to get up to altitude than in cruise, however it's close enough to illustrate the problem.
Typically extended range versions of an airplane add somewhere between 25-50% to the range, for example the King Air 360er you mention has 38% more range than a regular 360. You are proposing a 180% increase in range, which will almost always be outside the range of possibility.
One thing you also have to consider is it's not just about weight, you have to consider where that weight would go. Fuel weight has to be spread around center of balance, if you just add a bunch of fuel tanks at the back your airplane will be too tail heavy to fly. There's also the weight of the tanks, fuel lines and pumps you'd need to deliver the fuel to the engines.

The King Air 360ER isn't a good example, since the ER variant has a higher MTOW (16 500 lbs) compared to the regular 360 (15 000 lbs). If you increase the MTOW of the ATR, you could also get more range, but probably not that much...

So back of the envelope the best you think can be done by reducing the empty weight with an executive fit out, reducing passengers to 6 to 10 (instead of 42), and using empty weight savings as fuel is a 40% increase to 1000nm.  Still curious how the increased percentage of takeoff weight as fuel would extend the range as (like you noted) it is not a linear relationship.

I'm not saying it's a 40% increase in range @Edmund, that's the increase from the example you gave. A typical seat weights 15kg, getting rid of 30 seats only saves you 450kg, that's a 10% increase in range using the figures in my answer.

Yes, but you are forgetting the human weight. 42 humans at 80kg is 3360kg. Plus thier baggage say at 10kg each is another 420kg. So all up seats 450kg + humans 3360kg + baggage 420kg == 4230kg. Even if the executive furniture comes in at half the weight of the commercial seats we are still taking 4000kg of additional fuel on a lighter net fuel weight.

Your math doesn't really add up @Edmund, you may need to accept your idea isn't going to work.

@GdD Are you saying I'm unable to successfully perform a series of addition and subtraction? Please point out where I have gone astray in my demonstration that an additional 4000kg of fuel (basically doubling fuel capacity) can be added within the weight limitations of the aircraft by reducing passengers + baggage + furniture from 42 to 6 (max 10).

The empty weight of the airplane is without passengers and cargo @Edmund. The difference between the empty weight and MTOW is 7050kg. If you gave all that over to fuel, assuming that the weight loss from removing seats and the weight gain of the extra fuel tanks and piping equal out, then the increase in fuel is 55%, and that's with no seats, cargo or passengers. I've clearly articulated it in the answer and comments, take it or leave it, that's all I have to say.

@GdD Me'thinks I'll investigate just a bit further before I reach my conclusion. Thanks for the info thus far. I'll let you know.

I'd start with a cost estimate to double fuel capacity.  The plumbing mods won't be cheap, it will need an STC, and would likely require a large internal tank right next to your executive recliner, putting a damper on the ambience and likely busting your budget.

You could get a little closer with an ATR 42-600F conversion.
What you must do is figure how much the customized interior with passengers and luggage will weigh, the rest is fuel.
More weight could be borne with a larger wing with some sacrifice of cruise speed performance.  This would be off-set by the ability to carry much more fuel and ...
More power will give the ability to fly higher, essential to maximizing range$^1$ and improving TAS.
The ATR 42-600 is designed and highly specialized for the short-medium range market niche, whereas a long range aircraft, such as a Gulfstream G700 is needed here.
A lot would depend on how much customizing they would be willing to do.
$^1$ larger wing helps here too, allowing lower IAS for a given TAS, or Mach number.

Good idea to start bare bones and build up. It's all ready at 900nm fully laden. Wonder what it does empty. Could see if Comlux would do the outfitting.

I have fond memories of the Handley Page HP 42 Nothing in engineering is too ugly if it works well.

Good suggestion, but maybe better as a comment?

The 1940s solution:  two more engines, better supercharger, oxygen masks


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Along Track Distance ATD
ATD (Along-Track Distance) is used by ATC for certain types of (rare) NON-Radar Separation. I am training as an Air Traffic Controller in the USA. I have been told the only way to know if an aircraft is ATD is to ask the pilot. Once it is determined that the aircraft is using/capable of ATD, I can ask their distance from a fix, and calculate mileage needed achieve separation. FAA Publications clearly state that RNAV (Area Navigation) equipped aircraft have ATD, however, my question is: Can I assume that any aircraft with GNSS/GPS capability is able ATD? My guess is yes. I am unable to find anything that really ties this down. Thank you.

Did you ask your ATC trainer?


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How to choose an attack angle?
I'm writing a little piece of code for preliminary design that can calculate the ideal wing area for a given load, cruise speed and aspect ratio or span. It seemingly works well on the A380 data:
# A380
CONSTRAINS
Load: 475000
Cruise Speed: 907.48
Span: 80
Aspect Ratio: 7.5

Output:
GEOMETRY
Airfoil Lift Coef: 0.20400000000006546
Total Mass: 613525.1510617425
Lift Area: 851.9381984117002 m²               # <--- perfect
Average Chord: 10.666666666666666 m

However, I have a little issue relating to the attack angle with small airplanes:
# Cessna
CONSTRAINS
Load: 1157
Cruise Speed: 226
Span: 16
Aspect Ratio: 7.3

ERROR: Lift Coefficient was re-adjusted: 0.8950000000000661

GEOMETRY
Airfoil Lift Coef: 0.8950000000000661
Total Mass: 6796.908900200735
Lift Area: 34.68578659410046 m²                # <-- wrong supposed to be 16
Average Chord: 2.191780821917808 m             # <-- wrong

Now, we know the Cessna has a large attack angle, which increases the Lift coefficient to 1 and enables short distance takeoff capabilities at the cost of thrust. This incidentally reduces the Lift Area to 16m².
However, the same happens with the Piaggio and there is no explanation there.
My question is: How is the lift_coef and attack_angle chosen? There seems to be loads of variables at play.
Is it because the average chord is higher than what the wing can handle structurally?

Smaller, slower aircraft have different shaped airfoils, with higher lift coefficients, than a supercritical.  Recommend checking out the Cessna airfoil at airfoiltools.com. You will find good data on best AoA and its corresponding lift coefficient.  As a general rule, you pick the best AoA for cruising, and size the wing based on the weight it needs to lift at cruise speed.  Slats and flaps change lift coefficient at lower speeds.

Re "Is it because the average chord is higher than what the wing can handle structurally?" -- I'm struggling to understand the meaning of this sentence.  Are you suggesting that the wing would fail if exposed to the max lift coefficient?  If so, at what airspeed are you envisioning this happening?  (Certainly virtually any wing would fail if exposed to max Cl at redline airspeed!). Your question seems to lack enough constraints to make it meaningful.

Remember that the designer isn’t just optimizing for lift, they are optimizing for lift vs. drag.  Deriving lift by angle of attack is more costly in terms of drag than deriving lift by means of Bernoulli.  If I have a 3,100 pound plane designed to cruise at 150kts, then I want to design the airfoil so that it produces 3,100 lbs of lift from Bernoulli at 150kts, and as little as possible from AoA.  As the plane slows, the “Bernoullian lift” decreases and I have to rely more on AoA to make up for it, at the cost of increasing drag.

This is a rather simple question if you look at it from a slightly different angle, mind the pun.
If we assume cruise condition, that equates to straight and level flight. Here you know that:
$$ L = W = mg $$
By knowing the mass of the aircraft (m) you have the lift force (L), given g is a constant (9.81m/s2).
You need to know your cruise altitude, which then gives you your density (p = rho). You can always look this up in a table for ISA (the international standard atmosphere).
Hopefully where you got your other aircraft parameters, you also got a cruise speed.
Looking at the equation for lift,
$$ L = \frac{1}{2} \times p \times v^2 \times C_l \times S $$
you have everything except Cl, which you can now calculate.
You can then use thin aerofoil theory to give you the corresponding angle of attack needed for this. If using radians, thin aerofoil theory gives the lift slope for a Cl v AoA plot as 2pi. In degrees this is approximately 0.11 "Cls per degree". If you don't want to assume a symmetric aerofoil, you can assume between 0.2 to 0.4 as the Cl at AoA = 0. Alternatively, if you have knowledge of what the aerofoil is, you can consult an online database of aerofoils. This should give you the lift slope and the coefficient of lift when AoA = 0.
If you want, you can also correct for the AR. That is, the lift for a wing is less than the lift that should be generated by the constituent aerofoil. That is, the angle of attack in 3D flow needs to be greater than for 2D flow. For this you need the correction equation, and you need to know the AoA for Cl = 0. Happy to provide a link to the equation if you feel you need to go that far.
Hope that helps.

Welcome to aviation.stackexchange!


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When using momentum theory for estimating power needed to hover, is propulsive efficiency accounted for?
I am using momentum theory to estimate power needed for an eVTOL. I have read past papers where the results got from momentum theory was direcly use to estimate the power needed from the battery. How it that possible?
Should we account for drivetrain and propulsion efficiency (and divide the result got from momentum result by the efficiency to find the real power output needed from the battery)?

Yes we should, and fortunately there have been measurements to check how the momentum theory stacks up with reality.

The pic above was also used in this answer and shows the methodology for helicopters, where an extra loss factor of 1.15 seems to generate a good match. This answer provides some more details on computation of thrust from ducted fans, which I presume the eVTOL has.


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Power calculation of propeller
This is probably a very simple question, but I have been really struggling and need your help.
I have simulated a propeller with these parameters on Ansys:
Thrust: 2000 N
Angular Velocity: 314 rad/s
Torque: 200 N.m
Airspeed: 100 m/s
Propeller Diameter: 1.2 m
I got the following results (rounded for simplicity):
Thrust: 2000 N.
Torque: 200 N.m
Now I need the to find the efficiency, and I have used the formula Thrust*Airspeed / ShaftPower
For the shaft power, I have used Torque * angular velocity.
Using my numbers, I get an efficiency of 3.18, so there is a mistake somewhere but I cannot understand where. Are my parameters relevant? Could there be an error in the simulation ? Is it possible for 200 Nm torque, 1.2 meter diameter propeller to produce a 2000N thrust? Is my method of calculating shaft power wrong ?
Thanks

Given the basic information provided, a momentum assessment of efficiency of this propeller shows it to be about 4 or 5 percent too high, compared with a typically expected maximum of about 88 or 89%.  The error noted herein, and that by OP, may require accounting for induced flow. And shaft power, of what: is this prop 2 blades, or perhaps 3 blades? Power input is thrust times air-flow velocity through the propeller disk; power output is thrust times translation velocity of the propeller. That's easily calculated. More information about this propeller, & its application, would be helpful.

The equation Thrust * Airspeed / (Torque * AngularVelocity) is indeed the correct one for propeller efficiency, implying that we get an efficiency of zero at take-off.

For aeroplane propellers there are a few diagrams published, such as the one above from Torenbeek section 6.3 on characteristics of propellers: dimensionless power coefficient $C_P$ against J = airspeed/propTipSpeed. Listing efficiency, which is of course always less than 1.

For your question, a graph plotting thrust against power would be more interesting, and these graphs are mostly found in the helicopter literature, such as referred to in this answer. We can use this as an order-of-magnitude check for your derived parameters. The three equations valid in the hover would equate to propeller data at takeoff:
$$T = C_T \cdot \rho A (\Omega R)^2$$
$$Q = C_Q \cdot \rho A (\Omega R)^2  R$$
$$P = C_P \cdot \rho A (\Omega R)^3$$

Thrust: 2000 N Angular Velocity: 314 rad/s Torque: 200 N.m Airspeed: 100 m/s Propeller Diameter: 1.2 m

$C_T = \frac{T}{\rho A (\Omega R)^2}$ = 2000/(1.225 * $\pi$ * 0.6$^2$ * (314 * 0.6)$^2$) = 0.0407. Which is off the scale! We could be an order of magnitude off here.
What I would do is check all parameters and outcomes using graphs like above as a cross reference.

+1 for calcs.  We can see why James Watt analyzed the first meter of motion, before drag from air became significant.  The same can be done here, a measurement of acceleration (same mass, same fuel consumption).  Or one can essentially put different props on the exact same airplane and do a full scale velocity vs prop, holding aircraft coefficient of drag and fuel consumption constant.

Simply put, yes you are wrong somewhere, most likely in how you put together your Ansys simulation.
Thrust times airspeed gives power, and so does angular velocity times torque. You somehow get more power out than you put in, which doesn't make sense.
Ask yourself:

Is the thrust vector and the torque vector in the direction you expect? Perhaps you're simulating a windmilling prop?
Is the 100m/s freestream velocity or induced velocity?
How do my numbers compare to analytical equations for this problem?

All numerical packages suffer from garbage in/garbage out: if you give it nonsensical input, it will happily give you nonsensical output without any complaints. More details on your simulation would be required to exactly pinpoint where you're going wrong.

@Robert No. The OP clearly has an impossible situation where output power is greater than input power. There's most definitely a direct relation between shaft power and power "state" as you call it. I'm not going to remove my downvote since it shows a clearly lacking understanding of basic physics. Thanks.

@RobertDiGiovanni Power is conserved across mechanical advantage (unlike force, torque and displacement, rotation respectively). So 2kN acting on 100m/s velocity requires 200kW of power, regardless of any mechanical advantage. A rotation of 314rad/s and 200Nm produces ~36kW, regardless of any mechanical advantage. Power input is less than power output; clearly impossible. Please stop being so stubborn and learn the basics first.

First, we can convert 2000 N thrust to pounds thrust:

2000 N × .225 lbs/N = 500 lbf thrust

Next we convert rad/sec to RPM:

314 rad/sec × 57 degrees/rad × 60 sec/min × 1 revolution/360 degrees = 2983 RPM

Looking good so far.
Now we convert 2983 RPM and 200 Nm torque to Horsepower:
200 Nm × 2983 RPM/7127 = 84 Horsepower
These results are certainly near the range of reality, especially for a static bench test of a 1.2 meter propeller.  Thrust will drop as airspeed increases, which is one of the reasons why prop engines are rated in shaft Horsepower.  Thrust will vary dependent on pitch, RPM, and airspeed.
Engine Nm torque for a given RPM is the same as the angular drag torque of the prop at the same RPM.
Airspeed:  100 m/s. Converting to miles per hour:

100 m/s × 1 mile/1600 m × 3600 s/hour = 225 mph

This is a bit optimistic for 84 horses, but with reduction of weight and drag, and a variable pitch prop, one could get a bit closer to 225 mph than a Cessna 152, perhaps with something like this.
Regarding the prop efficiency calculation, it would seem far more meaningful to compare fuel consumption vs thrust against  an agreed upon "standard propeller" at a given airspeed and RPM.  Fuel consumption is directly proportional to prop drag.  Efficiency is (obviously) thrust vs drag of the prop, not the aircraft.

The OP posted everything in SI units (as one always should), and the first thing you do is convert to "pounds thrust" (for which the correct symbol is lbf, not lbs, by the way). Why?

@Bianfable Because it can then be compared with known values to see if OP is "in the ballpark" with the numbers.  I recollect around 3 lbs of thrust per HP at speed for props.  Please correct me if I'm wrong.


text:
Two low flying large planes made bank turn - not on radar
On Thursday March 10, 2022, at approx. 7:30pm in Monroe, Connecticut, I witnessed two full sized planes following one another extremely low and then make an incredible bank turn.
These planes did not show up on radar apps/websites. They came up from the South and banked South East, most likely returning in the same direction they came from.
I have never seen a plane that low be so quiet…and then came another one right on its tail. My guess is they were military or some sort of test run from Sykorsky Memorial Airport.
If anyone has any info on how to find out what happened, I would really appreciate it.
It’s been driving me nuts trying to figure out what kind of planes they were, why they were so low, why they both made a crazy U-turn at that altitude…so many questions have been racing in my mind.

Start carrying a camera in case they come back.

Did you take notice of how many engines these aircraft had? Were they low wing or high wing design?

"The planes did not show up on flight tracking websites" and "The planes did not appear on radar" are extremely different statements, and you have no way of saying whether the second statement is true. And it's possible that they would have appeared on ADSBexchange, which very prominently does not block military aircraft, even if they weren't on FA or FR24, which do block.

What does “full-size” mean, other than “not a drone”? How far away were the aircraft in question?

If you feel like the Hail Mary, you can always ask the FAA what's up?  They explicitly can't tell you what's going on with military aircraft but if there's a base near you, the FAA suggests you contact that base's noise complaint office.
Probably get nothing, but the worst it'll cost you is an hour of your time.
I'm not aware of Sikorsky seeing much military use, but KCEF's an Air National Guard base not that far to the north.  There's no military use areas on any charts I have for that area.
My money, though, says that if they were super quiet, it's possible they were gliders? There's a glider ops area about halfway between Danbury (DXR) and Sikorsky (BDR) which is roughly where Monroe is, depending on what part. They've got huge wingspan, would be basically silent and while I'm skeptical that you'd make a hard bank turn while under aerotow, they fly pretty slow so I expect their turning radii are pretty tight. It wouldn't shock me at all to learn that two glider pilots decided to fly together for funsies, especially that close to a national park. I bet the views are amazing.


text:
Are these units correct for Hp/W
using the table provided, and how it says the Velocity should be in mph, is it correct to say that the units of Hp/W from this table is mph also?

Some context for where this table comes from might help you get an answer. This may also be better suited to Physics.SE or Engineering.SE.

This is from Aircraft Design: A Conceptual Approach by Daniel Raymer, as @Raketenolli guessed. (page 81)

This is an empirical formula showing a simple relationship observed between power-to-weight ratio and maximum speed for different types of airplane. Its source is most likely an aircraft design book by Raymer (it does look familiar!) or Roskam.
For the formula to work, all physical quantities have to be made dimensionless, because you can't really raise a speed to a non-integer power (and even for integer powers, it only makes sense in a few cases). So you have to know exactly which units were assumed for power (presumably hp = 746 W), weight or mass (lbf = 4.45 N, lbm = 0.454 kg) and speed (mph = 0.447 m/s).
An example:
Assume we are designing a metal homebuilt airplane that is supposed to fly 120 mph max. We check Row 2 of the diagram to find the proper values of $a$ and $C$, then plug those values into the given formula:
$$
\begin{align}
\frac PW &= a {V_{max}}^C \\
   &= 0.005 \cdot 120^{0.57} \\
   &= 0.0766
\end{align}
$$
Assuming horsepower and pounds and an MTOW of 1000 lb, we find our engine needs to have
$$
\begin{align}
P &= \frac PW \cdot W \\
  &= 0.0766\, \frac{\mathrm{hp}}{\mathrm{lb}} \cdot 1000\,\mathrm{lb} \\
  &= 77\,\mathrm{hp}
\end{align}
$$
which seems like a very reasonable value for a small 1000 lb airplane.
If you wanted to use different units, e.g. SI, then you would need to adapt the coefficients $a$ and $C$ from the table.

yes, it's from Raymer's book.

Yes. It would be correct to say that "power divided by weight" is a unit of speed. Given that the equivalent speed is given in miles per hour, the power-to-weight ratio will also be in miles per hour. Note that when I say "weight" here I really do mean "weight," that is, the force exerted due to gravity acting on the mass of an object; the colloquial definition of "power-to-weight ratio" really means "power-to-mass ratio," and that is not equivalent to a speed.
Using SI units, power divided by weight is given as Watts per Newton:
$$
\frac WN
= \frac {\mathrm{kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-3}} {\mathrm{kg} \cdot \mathrm m \cdot \mathrm s ^{−2}}
= \frac {\mathrm m}{\mathrm s}
$$
which is sufficient to convince us that "power-to-weight ratio" is nothing more than a complicated way of expressing a speed, in particular the resultant speed of an object of some given weight when it is pushed by some given power.
In the English units here, "power" is in horsepower and "weight" can be assumed to be pounds-force:
$$
\begin{align}
1 \, \frac {\mathrm{hp}} {\mathrm{lbf}}
 &= \frac {550 \mathrm{ft} \cdot \mathrm{lbf} \cdot \mathrm s^{-1}} {\mathrm{lbf}} \\
 &= 550 \, \frac{\mathrm{ft}}{\mathrm s} \times \left( \frac {1 \, \mathrm{mile}} {5280 \, \mathrm{ft}} \right) \left( \frac {3600 \, \mathrm{s}} {1 \, \mathrm{hr}} \right) \\
 &= 375 \, \mathrm{mph}
\end{align}
$$
So we expect that an object which weighs one pound and is acted upon by a force of one horsepower will equalize at a speed of 375 miles per hour. But! the power given in the equation is not the true force acting on the object; rather it is the power generated by the power plant, which cannot be 100% efficient. Also, our 375 mph figure assumed no air resistance; in fact air resistance does exist, and it is a complicated topic, but generally (for high-speed flow) increases with the square of the object's speed. So when your textbook says
$$
\frac{\mathrm{hp}}{W_0} = a V^C
$$
they really mean
$$
V_\mathrm I = a V_\mathrm R ^C
$$
where $V_\mathrm I$ represents the "ideal" speed in a vacuum and using a power plant whose energy output is completely translated to speed with no losses along the way, and $V_\mathrm R$ represents the actual speed you will see in the real world.


text:
Converting 2d Cl to 3d
I am trying to convert 2d Cl data obtained from javafoil for a NACA 4412 into 3d data to be used for analysis of the airfoil, but cant find any equations online

It's unclear what you're asking. By "convert into 3D data" you mean you want to determine the lift or lift distribution of an entire wing that uses the NACA 4412 airfoil? Because the airfoil by definition is only two-dimensional.

3D data, as in the third dimension is the wing span: the performance of the whole wing is the integral of the wing sections from root to tip. For the local 2D lift l of the section: $$\frac{dL}{db} = c_l \cdot \frac{1}{2} \rho V^2 \cdot c$$ with $c$ = chord length of the profile and $b$ the wing span. For the complete wing:
$$L = \int_{root}^{tip} c_l \cdot \frac{1}{2} \rho V^2 \cdot c \cdot db = C_L \frac{1}{2} \rho V^2 \cdot S$$ with $C_L$ the average lift coefficient, S the wing area.

Local $c_l$ is a function of local angle of attack $\alpha$. Along the integral, the local $\alpha$ and therefore the local $c_l$ will vary because of:

Wing twist (washout).
Interference effects from the fuselage
Vortex generation around the wing tip

with the last two effects most noticeable on swept wings, as depicted in fig 7-19 of Torenbeek. Also, the local chord $c$ changes due to the taper ratio of the wing, and the local stream velocity V will change as a function of interference with other bits of the aeroplane. So in order to compute the 3D wing data, the wing configuration must be known. The integral can be solved mathematically if the functions of $c_l$ and $c$ and V along the wing span are known.
Even then there are issues with the local streamline fields that are hard to capture in an equation, and comparison with statistical data of existing aircraft wings is useful. So are Computed Fluid Dynamics and wind tunnel measurements


text:
Oswalds efficiency factor for trapezoidal wing
For a part of my university project we have to approximate 2d to 3d data ideally using oswalds efficiency factor, the equations given to us are for either rectangular or swept wings however the wing I am using is a trapeizoidal wing as seen below and so I am unsure of which approximation to use. Any help would be greatly appreciated

I'd use the average value from the first two equations, it's anyway just a very rough estimation.

It looks like the wing of the Fokker 27, you should be able to find the Oswald factor for that one.

what not present calculations for all three?
First, some background:
$C_{drag}$ = $Cd_0 + \frac{(Clift)^2}{\pi\varepsilon AR}$
A smaller $\varepsilon$ means a larger coefficient of drag, and, right there, is the AR (aspect ratio) contribution to the formula.
Also, one can notice the (Clift)$^2$ term, explaining the shape of the drag vs AoA curve.
OK, AR is around 7.5.  3rd calculation first:
$\varepsilon$ = 1.  The famous Spitfire wing, beautifully tapered leading and trailing edges, harder to build
Treated as a rectangular wing using the first calculation:
$\varepsilon$ = 1.78 × (1-0.045(7.5)$^{0.68}$) - 0.64 = 0.825
Treated as a swept wing with 0 sweep using the second calculation:
$\varepsilon$ = 2/(2-7.5 + $\sqrt{(4+7.5^2}$) = 0.885
What did the Wright Brothers find with their wind tunnel?

Rounding the trailing edge wing tips reduces drag

Half a Spitfire wing.
Without going any further, as we do have a policy regarding homework, one may lean toward the second equation ... BUT

why not get wind tunnel data for your wing for comparison?

This may elevate your work above and beyond an ordinary "assignment" to one worthy of "post graduate"$^1$ work. Good luck!
$^1$ something an aerospace company may be interested in


text:
what exactly Vx and Vy for in gliding and powered descent(piston-prop aircraft)
as i understood,for a prop-piston aricraft :
during a gliding, through the equation:
tanγ= D/W≈ 1/K
sink rate≈ VD/W≈ V/K
by which we get(theoretically):
Vx(Vmp) is the speed at which we get minimum sink rate
and Vy(Vmd) match up with minimum γ
but when it comes to a powered descent:
tanγ= △D/W
sink rate≈ V△D/W= -△P/W
then because at Vx(Vmp) the prop aircraft gets minimum excess drag gets minimum γ
at Vy(Vmd) gets minimum -excess power gets minimum sink rate
which does not add up, can't even make a
graphic for powered descent.
what is wrong with that?

This looks like a sort of homework question, which is fine but it's not entirely clear what it is you're actually asking - "What is wrong with that?" is a fairly broad problem statement without much focus.

Vx (the speed of the steepest climb, ie. speed with most excess force) and Vy (the speed of the fastest climb, ie. speed with most excess power) are defined with climb power.
I don't really see they should play any role in descent, let alone in glide in which by definition there is no engine power. Vx and Vy are tightly defined by the engine properties and especially in a piston-prop where the engine thrust varies greatly with airspeed. They should not be mixed with by Vmp and Vmd even if they coincide.

This might be easier conceptually to look at thrust as negative drag.
A powered descent means a flatter angle of descent than an unpowered descent, you've just turned your Cessna 172 into a high performance glider!
So, to get maximum distance, you still want to be around Vbg to have optimal wing AoA for maximum Lift/to Drag.
Next you want to see at what speed your prop gives best efficiency at its AoA.  Keep in mind a powered descent will be lower RPM than full throttle climbing Vy, so (for a fixed pitch) best prop efficiency will be at a lower airspeed.
The descent "Vy" will be near Vbg and V max prop efficiency.  As far a descending "Vx", one may consider flaps (more drag) and no power for a steeper descent angle .


text:
Why the greater the propeller slip is, the more thrust it generates?
Forgive my poor English skill and my entry-level understanding of aviation stuffs. I'm currently a student pilot.
As I understood, the propeller slip kind of represents the propeller efficiency in some way, and when slip gets greater, the more thrust is "wasted", in another word, they are not being used on moving the aircraft forward. So why they are doing that way? It's confusing me.

Hi there. What is your source for this information.

@Jpe61  It's on my textbook, and chinese pilot license exam as well.(I'm a chinese btw).  Now I'm not sure if this whole concept is correct XD.

Well, your english is just fine, do not worry about that. As I have not seen this claim myself (or just can't remember, my readings on prop theory are ancient), I can only speculate: I think this is because at high power settings the  the propeller simply loses more and more "grip". While the propellers does create more thrust, it is highly inefficient in this condition. There are two ways to produce high thrust: accelerate a little air a lot, or accelerate a lot of air a little. Propellers use the latter method.

It makes sense, Thanks a lot! I finally get a grasp on this :D @Jpe61

This is actually a very good question in that it can be interpereted in two ways for aircraft:

The acceleration point of view

The steady state "cruising" point of view.

From a standing start, increasing rpm means more turns per meter.
Technically, until the plane speeds up, it is "slipping" more and producing more thrust.
When cruising at constant speed, you want as little "slip" as possible. In other words, lowest rpm per unit thrust.  This, of course, is a function of the propeller being at its best AoA for greatest efficiency when peak thrust is not needed.
This is why complex aircraft "flatten" their prop blades for takeoff and reduce rpm for cruising.

Got it! Thanks a lot!

Here is a good definition of "slip" jargon relative to propellers.

The concept that the propeller is "slipping" is just another way to express the idea that the propeller is meeting the air at a non-zero angle-of-attack.  It's logical that for a given speed of motion of the prop blade through the air, the greater the angle-of-attack-- i.e. the greater the slip-- the more the thrust, as long as the angle-of-attack is not so high that the prop blade is stalled.


text:
What is the maths behind Loitering of Aircraft?
I am making a controller for loitering using PIDs.
The equation which is used to find the radius of circle given bank angle and speed is in following link:
How to calculate angular velocity and radius of a turn?
What i did was give the speed and current bank angle of aircraft to find the radius it will make. By subtracting this radius value with the required radius of circle i got the difference between them. This difference was given to PID "Roll from Radius Error loop", this generated the required roll value which is given to "Aileron from Roll loop" and in this way the required circle radius was achieved. But there was a problem.
The Loiter is around a point but with above simulation the aircraft do not loiter around the required point. In other words the center of circle was shifted.
My question is how can i loiter around a point what is the math behind it or how can i loiter around a specific point.

What are you building/making the controller in (eg what language)? What controller inputs do you have? This sounds like a homework problem as written.

Programming drone logic is way outside the scope of this forum. Drones.SE might be a better fit, although the breadth of the question seems like a better fit to a chapter or two of a textbook, rather than the paragraphs of a Stack Exchange answer.

@RalphJ The question is perfectly valid for aeroplanes as well. The paragraphs of DeltaLima's answer seem to do the job just fine. Voting to keep open.

Did you account for wind?

@tedder42 i am making flight controller in MATLAB/Simulink

@StephenS yes the crosswindspeed will be feedforwarded to find roll.

What you have built is an open-loop controller. You do not feed the distance between the aircraft and the specific point back to the control loop. And thus if the aircraft deviates momentarily from it's intended radius of curvature, it will recapture the intended radius, but now the circle centre has moved.
You should program your controller to not only fly a certain radius, but also to fly a constant distance from the specific point.
On the ideal path:

the curvature radius is equal to the distance to the specific point
the speed vector is perpendicular to the vector between the aircraft and the specific point
the distance to the specific point is at the target value.

Now for the controller:

You control the radius of curvature with the bank angle
You control the speed vector (track angle) with the radius of curvature
You control the distance to the specific point with the speed vector.

The model will be non-linear, so for larger deviations from the target distance, a PID controller without any limiting logic may not be stable. You may need separate controller logic to bring the aircraft to an acceptable initial state before activating the constant turn controller.

thank you for the answer, you got my point. I will keep these things in mind and implement it.

@Asad I am glad I could help. Wishing you a clear mind, lot's of fun, a bit of luck (it seems one always needs a bit of luck for controller implementation) and finally success during the implementation.

In the multidimensional matrix of stability and control of aircraft, maintaining an earth reference position means unfortunately that the controller needs to consider multiple signals. Also, it needs direct feedback of the controlled variable - the GPS position. @StephenS asks a very valid question in a comment: “Did you account for wind?”
The referenced answer shows the math valid for an aeroplane in a co-ordinated turn: a quasi-static situation, all is trimmed such that altitude remains constant, no inter-dimensional accelerations such as the propeller torque changing the required bank angle. Or the dynamics for angular acceleration: it takes some time before bank angle is reached after deflecting the ailerons (and rudder, to maintain the turn being co-ordinated). And in that situation the turn radius is constant with regard to wind axes, not earth axes.
The controlled variables are GSM position amd altitude. The main input signals are airspeed and bank angle. If there is a direct feedback of distance to GSM position, the error in distance can be used to apply aileron/rudder/elevator deflection, engine power to fly upwind/downwind etc. a full simulation of all physics is required.

Generally in such a situation you want two nested controllers.
The "inner" controller adjusts the plane to achieve a desired bank angle.
The "outer" controller adjusts the bank angle to achieve a desired flight path. You can make small adjustments by alternating tighter and looser turns around the circle. To move the centre of the circle east fly looser curves when flying east than when flying west. For larger adjustments you probablly want to switch to a completely different navigation mode.

Yes exactly. I am following this scheme
"Roll from Heading Error"->"Roll rate from Roll error"->"Ail. from Roll rate error"


text:
Can I transition the airspace (pictured below) without 2-way communication?

If I wanted to fly over Beverly regional between 2600-3000ft, do I need to speak to anyone as I do so? I would be above the delta and below the bravo, but the sectional states I must ask for approval from approach (124.4) above 2500ft? If so, what airspace would that be?


text:
Under which circumstances were 727s or 737s used for long-haul flights?
I've been intrigued by the occasional picture of a 727 crossing the Atlantic, like this Air France in Miami from the late 1960s. Under which circumstances would such narrow-body aircrafts (most notably 727s and 737s, or even A319s and A320s) be used for long-haul, intercontinental flights?
PS: I'm referring to "regular" (e.g., flag) carriers, not private/cargo jets.

If you have what is called a long thin route, then a narrow body is ideally suited. The better example of this is the B757. The new A321XLR adds nearly 2000 km to the range of the A321, up to 8,700km! This will make transatlantic flights between smaller cities on both sides much more common. If it is not immediately obvious, a long thin route is a city pair route that is a relatively speaking long distance, and also does not have the demand for one or more daily B747 or A380 flights (thin = not much demand). It does not mean ultra long range, where an A350 or equivalent would be needed (project sunrise in Australia, for example). These could be both business or tourist routes.
I should also add that the B727-200 would not have been able to fly the distance from France to Miami non-stop. Looking into this, it is a flight from the French territories in the Caribbean (Pointe-à-Pitre).


text:
Why don't all twin engines counter-rotate?
It seems that having counter-rotating twin engines would be a natural choice to avoid the unwanted yaw seen in single engines and yet there are several twin-engine aircrafts that that aren't counter-rotating. Why is that?

Cost-savings.  It is easier to install the exact same model engine on the aircraft than two separate model engines.  It also allows owner / operators the ability to find and use a loaner engine easier than airplanes with counter-rotating props.  IMO.  I don't know how difficult it is to switch the direction of rotation on the engine.

Your assumption that only “several” twin engine aircraft don’t have counter rotating propellers is incorrect. MOST twin engine aircraft don’t have counter rotating propellers.

@MikeSowsun Do you know why most twin engine aircraft don't have counter-rotating propellers?

It is only cost savings because having one engine turning the opposite way complicates manufacture and maintenance

@Tfovid - re "Do you know why most twin engine aircraft don't have counter-rotating propellers?" - because they have no propellers?  Seriously though, nothing in body or title makes it clear the question is limited to aircraft with propellers, might want to consider an edit.

@quietflyer If nothing makes it clear that the question is limited to propellers, it's because it isn't limited to propellers. As for "Seriously though", consider using a better tone if you want to make SE a better place.

@Tfvoid -- tone is hard to detect in the text media. "Seriously though" was me laughing at myself for jokingly suggesting that you might have meant the question to encompass pure jets, not chastising you for posting a question that encompassed pure jets.  But so that actually was your intention, to encompass jets?  As sort of a trick question, or because you thought it would be obvious and l ogical to all that that's what was intended?

Engines are mounted slightly canted downward to eliminate the P-factor in cruise, and the rudder has to be sized to compensate for yaw due to asymmetrical thrust, the P-factor on take-off is tiny compared to that, so the pilot has no problem compensating it. And the pilot has to use the rudder for other reasons, like compensating for cross-wind, anyway.
The P-factor does play role in the engine-out situation, as it makes the outward-rotating engine produce a bit more yaw. But an outward-rotating engine also causes less rolling moment toward the other engine, see e.g. Why is an outward spinning propeller better in theory on the North American F-82 Twin Mustang?, which improves the engine-out handling in other ways. And it just takes making the rudder a bit bigger.
According to this answer to Do turbine engines on multi-engine aircraft rotate in opposite directions to offset torque?, the main reason for counter-rotating engines is actually stall characteristics. But still, that can be fixed by other means like appropriate wing twist.
On the other hand, having to stock two different types of propellers and gearboxes (the engine can spin the same way and have to rotation reversed in gearbox on one side), and the additional risk of mounting them wrong, are a huge complication to maintenance. Aircraft already cost a lot to maintain. So if a slightly bigger tail and wing twist make the design controllable enough, including with engine out, without having to deal with counter-rotating propellers, that's the design that gets chosen.

Continental used to make reverse-rotation versions of a few of their engines, specifically for light twins.  All that's required (assuming gear, rather than chain drive for the camshaft) is a different camshaft and starter motor (plus different propeller, of course).  Cost was barely any more than the standard rotation version (lower volume cam and starter production). Errors are a much bigger factor than cost in why more reverse rotation engines don't exist.

@ZeissIkon, note that I did not mention manufacturing. It is indeed mainly a maintenance problem.

In fact I will just do it; roll back if objectionable.

(Deleting lengthy explanation of tiny edit; see your inbox if wish to read it ;) )

(Or did you mean each engine was canted slightly to right (if props turn CW)?  If so, better edit my edit.  Sorry and thanks-- )


text:
How do I calculate the centre of gravity for the below question?

Textbook Answer for above question: BEM of 25,450kg & the CG is 57 cm in front of the main wheels.
I'm having trouble on finding the CG position. Is there any ways to find the datum of the aircraft in order to calculate arm, moment and centre of gravity position?

If the CG were above the main gear then the loading would be 24725 on the main and zero on the nose.  If the CG were directly over the nose wheel then there would be 24725 on the nose and zero on the main.  So the ratio is the loadings gives you the ratio of the distances between each wheel and the CG, in this case 24000:725.  By inspection we can see that the CG is closer to the mains than the nose wheel.  Is that enough to point you in the right direction?

At first, I don't quite get it, to be honest. But after reading it a couple of times, I think I get a brief idea. So based on my understanding, we can roughly assume the CG position by using the ratio method to determine where it is likely to be closer to either the nose wheel or mains wheel. Do I understand your explanation correctly? Thanks for your explanation, by the way.

having a formula is good but I’ve always preferred the idea of having an intuitive understanding of the problem, visualising what’s happening and then applying the numbers.  In this case, most of the weight is on the main wheels and a little on the nose wheel so it follows that the CG must be a little in front of the mains.  If you get a numerical answer that doesn’t agree then you know there’s a problem.

Putting the numbers in, the ratio 24000:1450 (sorry I counted the nose wheel loading incorrectly before) applied along the 10m distance gives 9.43:0.57metres so the CG is 57cm from the mains and 943cm from the nose wheel.

I agree. Well explained, and thanks again for your great help and clarification.

To find the position of the centre of gravity, you need to find the point where the sum of the moments of the weights on the wheels is 0.
$$\sum_{i=0}^n \left[W_i \times \left(x_i-x_{c.g.} \right)\right] = 0$$
You can choose the datum point yourself; the centre of gravity will be found with respect to that datum point.
It often helps to put the datum at one of the axles, because it simplifies the equations.
If we put the datum at the main wheel axle, the equation becomes:
$\left( 2\times 725 \right) \times \left(1000 - x_{c.g.} \right) - \left( 4\times 6000 \right) \times  x_{c.g.} =0 $
Solving for $x_{c.g.}$ is left up to the reader, but it should give the correct answer.

Hi DeltaLima, the formula does work. Thanks for your big help. I have been stucking on this question for quite a while.

Hi again, I'm not very familiar with the formula that you used because I'm quite new to the topic of the centre of gravity, so I don't have a strong foundation. Is there any recommended place where I can better understand how the formula works and practice?

@ChuaRenyu - In English, the formula says "the sum of all moments is zero" and "moment is weight times lever" and "lever is distance from c.g." Don't memorize the formula, learn the meaning.


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How is the hazard from a malfunctioned ejection seat system handled?
Suppose an incident like this:
An airplane equipped with an ejection seat suffers a problem severe enough that the pilot decides to eject. As he pulls the handle... nothing happens. The ejection system completely fails to function.
Upon realizing that he's either going to land the stricken plane or go down with it, the pilot somehow manages to maintain control and bring the aircraft into a successful landing ("successful" meaning non-fatal for the pilot, if not for the aircraft). On the ground, the emergency crew rushes to release the pilot from the cockpit.
Now, one would suppose that a failed ejection seat system presents both the ground crew and the pilot a similar kind of a hazard that a dud ammunition would: firing has been attempted but it has not commenced for an unknown reason, and the system with all the explosives in it is now in a potentially unstable state, unsafe to approach or access.
How is such a situation handled? Are there any standardized procedures to disarm and secure a failed ejection system from the outside as part of the rescue effort?
Have there been incidents like this in real life?

This depends on the design of the seat, which is not so much an "accessory" but a "deeply built in" part of the aircraft system.  A couple of things...  The pilot normally stows their own seat safety pins onboard prior to taxi, so the person "safetying" the seat on landing will be the pilot.  Remember also that the ejection is a sequence of several events including canopy ejection or breaching, limb pre-tensioners, etc.  So the pilot does know how far into the sequence the seat got before failing.  This leads to the appropriate checklist upon landing for both the pilot and ground crew.

@Max, maybe, but probably not.  The sequence is extremely fast.  And if the seat fails 99% of the time time pilot will be dead.

@Max: at what point in the sequence does the limb tensioning happen? I'd imagine it is pretty early, so essentially, as soon as the pilot pulls the handle, they will be "handcuffed" to the seat, unable to reach the flight controls, will they not? This whole scenario seems extremely contrived. Both the aircraft and the seat are so critically damaged that the pilot choses to eject and the ejection fails, but then magically the aircraft becomes completely fixed and flyable again?

@JörgWMittag These are interesting questions, a lot of things are happening in sequence, many of which are external to the seat, and I don't know the specific sequence.  Even more complicated is sequenced multiple crew ejections.

Chain of events:
Aircraft in flight
Becomes uncontrollable
Pilot eventually comes to the conclusion of "I have to get out"
EJECT EJECT EJECT (pulls the relevant handles)
The ejection fails (is the canopy still on?)
Aircraft magically becomes controllable.
Pilot declares IFE (In Flight Emergency)
Pilot lands.
I expect it would be met with all manner of emergency personnel.
Crewchief, Asst crewchief. Fire, EOD, WPNS ...
First order of business, get the pilot out of the seat.
Then, carefully safe the seat (reinsert pins) or whatever is needed for this singularly weird condition that has probably never happened before.
After Crewchief, life support, EOD, possibly Weapons all get to the point of "safe"...take the thing apart and start the major investigation in conjunction with MartinBaker.

The aircrew have their own pins in their possession in the aircraft.  Many cockpits have a dedicated pin stowage pouch.  The pins MUST be inserted by the pilot before he/she gets out of the seat, not after.

@Max - In a normal 'getting out of the aircraft' sequence, sure. Pilot puts the pins an and then gets out. But, after a failed ejection sequence? Not so sure that would be a "MUST" condition.

You can't pin the seat while you are sitting in it, you can only safe the top and bottom handles.  There are 7 or so pins in all kinds of different places, all impossible to reach while sitting.

@WPNSGuy I am not an expert on this, I have been around ejection seats but have never maintained or operated them.  Michael may have more domain experience here. My point is that the pilot is in their "point of maximum safety" while still strapped in the seat.  If the seat motor fires on the ground, then they ride the seat up and deploy, with all the same risks of any ground ejection.  But once they unstrap, they now move into the situation of "infinite risk," if the seat fires they will have little chance of survival.  I'd stay strapped in ALL DAY until it was "safed."


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What is the cause of possible crash in an accidental low RPM takeoff?
I've read from quite a few sources that during the World War II, it was not that uncommon a mishap for a fighter pilot to take off with a "coarse pitch" / "low RPM" prop setting. This would typically happen as a result of pilot fatigue after a prolonged period of intense activity, aggaravated by the fact that in a single-seat fighter there were no other crewmembers to warn about the incorrect setting.
Sometimes, though apparently not always, the result was a crash. Even if there was no accident, writers depicting these incidents would usually use a tone indicating that they considered such a takeoff a first-rate blunder and an indication that the squadron as a whole was overtaxed and in an urgent need of rest. So, clearly it was a serious pilot error.
I started to think about what would go wrong in such a takeoff and found out that it's not at all obvious to me. I know how constant speed propellers operate, including the analogy of blade pitch with automobile gears, and understand that the engine isn't able to provide maximum performance on a low RPM setting, even at full throttle. However, since there were successful takeoffs as well, and especially because the lack of performance seemingly wasn't so severe as to make the pilot realize that something was wrong and to abort the takeoff run, it doesn't look to me like the reduced power alone would readily explain the mechanics of a crash in that situation.
So, what was the fundamental reason for a low RPM takeoff to end up in a crash? Did the aircraft simply roll past the runway's end or stall upon rotation due to the lack of power? Or would the engine fail catastrophically because of misuse? Surely the pilot would see from the instruments that airspeed wasn't building up as intended if it was just about a deficit of power?

There are two effects from taking off with the propeller levers at MIN RPM.
Firstly, horsepower is limited by the propeller governor, preventing the engine from developing takeoff RPM. It's not unlike a car with an automatic transmission stuck in high gear. Acceleration will be sluggish, and if you're on a hill with a trailer and the car needs to downshift but it can't, you're going to bog down on the hill.
So a pilot taking off with MIN RPM set will simply have a pretty severe performance penalty, with the engine only able to make the HP it can make at redline Manifold Pressure at minimum RPM.  A supercharged radial making 2000 HP at a takeoff RPM of 2700 might make only, say, 1100 hp if it's limited by the prop governor to, say, 1600 RPM (or whatever the MIN RPM setting on the prop control commands).
If you're flying a Corsair fully loaded with ordinance, and the engine is limited to 1100 or 1200 HP, you're probably going into the weeds at the end of the runway unless it is very long. Even if it gets airborne, it'll barely be able to climb.
The other issue is blowing up the engine itself. There is a value called Brake Mean Effective Pressure (BMEP) that is the total peak pressure in the combustion chamber, and this peak pressure, for a given throttle opening, is higher if the engine is turning slower. BMEP is a factor in the onset of detonation.
Most naturally aspirated engines can tolerate high manifold pressure and RPM below normal values (the pressure being limited to atmospheric), but when you add supercharging or turbocharging, the stress levels on the cylinders go way up.  So forcing an engine to run with 20 inches of boost, that is, 45" or 50" inches of manifold pressure, plus the engine held down at minimum RPM by the prop, can raise BMEP levels out of spec, causing detonation, and you take the risk of blowing cylinders up. So you end up in the weeds with ruined engine to boot.

This looks like a great answer! However, there's the other pilot aspect I'd like to know: why wouldn't a pilot abort the takeoff if the aircraft is clearly not speeding up like it should? Pilots involved in these incidents were often not newcomers, so I'd assume they already had developed a "gut feeling" about how the plane would accelerate in a normal takeoff, and also that a fully loaded fighter-bomber would speed up very sluggishly if the engine was only making roughly half of the usual takeoff power. Was the slow acceleration somehow evident only when it was already too late to stop?

You would have to have someone that is nearly catatonic from exhaustion to do this in the first place. The throttle and prop levers are adjacent to each other on the throttle quadrant, and you normally confirm with your hand that both are forward as you set takeoff power with throttle.  So someone that burned out could miss just about anything, including other, fairly obvious, cues that something is amiss..

A factor not directly related to performance (takeoff roll, climb rate etc.) is the torque exerted on the airplane on the takeoff roll. WW2 fighter planes with their huge engines and propellers are a handful to keep on the centerline even when operated correctly, the increased torque caused by the coarse prop pitch will make the problem worse.
One of the leading causes of accidents for the Bf 109 was groundloops on the takeoff roll.

My airplane has a constant-speed prop, and here's where I suspect the problems arise--and as has been suggested a lot of it is related to pilot preoccupation, pilot fatigue, or a failure of a propellor governor.

On a hot day in something like a fully-loaded, fully-fueled heavy bomber, you may absolutely need 100% of the airplane's performance to make it off the runway that's available.  So operating at 90% or even 80% of peak horsepower could absolutely put you into the trees.  It doesn't seem like the prop pitch would cause much of a hit, but to use your gear shift analogy, imagine trying to drive a manual pickup truck uphill, but starting in 3rd or 4th gear.  It can't be done.  Now imagine a load of bricks in the back.  It double-super can't be done.  The nature of absolutely global-scale combat operations often involves having to operate on the absolute fringes of the performance envelope in terms of loaded weights, fuel loads, shortness of runways...  and urgency to proceed.

Imagine I have a large displacement, supercharged engine making the maximum allowable cylinder pressure.  Oh--and the prop is creating a very large torque on the crankshaft because the coarse, low-RPM pitch directly resists engine torque.  Now I have a huge problem, I could easily over-temp or over-pressure a cylinder to failure before I even get off the runway,  My 4 engine, max weight, hot-day B-17 is now a 3 engine, max weight, hot-day B-17 and an accident is unavoidable.  In many GA planes, it perfectly acceptable to fly a little over-square, like 24" MP into a prop governed at 2100 RPM.  But in the engines of the 1940s, PLUS huge displacements, PLUS supercharging, you can be in a world of trouble if you aren't managing MP in concert with RPM.


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Updating KLN 900 Unit database
I have been trying to update KLN 900 database using the usb cable however getting the msg "Data Loader not ready" on KLN unit and on laptop screen "Not responding" with data-loading not moving past 0%. I have tried to trouble shoot as per the guidelines of Honeywell, still no joy. Can anyone help through this. COM port 4 is being used.

This is off-topic. I don't see anything related to aircraft or aviation in your question. Maybe try SuperUser.SE?

@AircraftEnthusiast007: This is not off-topic, it's about a GNSS based instrument providing aeronautical information. Whether it can be answered without asking for additional details is another question.


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How do I gain a PPL for small crafts such as a C172?
I'm heavily inclined towards rocketry and aviation, although I do not plan of pursuing aviation in my future (but instead rocketry) I'd love to be able to fly small aircrafts such as the C172 or a small Piper. As of now I'm only 14, but I'd like to know what future qualifications I might require in order to operate a C172 and what I'd need to study. (I do understand that this requires quite a budget).

It varies a lot by your country.

Depending on your location there may also be a lot cheaper options than PPL license. Are you interested in hearing about these?

This is easily answerable by Google, but also by visiting your local flight school.

I'm answering for how it is in India because I think from your profile that you're in India.

First Steps
You can start flying training at 18 years old for a Student Pilot License (SPL). To get a full Private Pilot License (PPL), you also need to be 18.
Qualifications and Training
You must have completed 10th standard (secondary school) at a minimum.
A 12th-grade education (higher secondary) with physics and mathematics is required for pursuing a Commercial Pilot License (CPL).
Student Pilot License (SPL)
First step towards becoming a pilot.
You’ll undergo basic training and take a few written exams. The SPL allows you to fly under the supervision of a certified flight instructor.
Requirements:

Pass a medical examination (Class 2 Medical by DGCA).
Clear a basic written test covering aviation knowledge.

Private Pilot License (PPL)
After obtaining your SPL, you can start working toward a PPL, which allows you to fly solo or carry passengers (but not for commercial purposes).
Requirements:

At least 40 to 60 hours of flying time, which includes dual flight
instruction (with an instructor) and solo flight time.
Ground school courses covering subjects like air regulations,
meteorology, air navigation, and aircraft technical knowledge.
Clear written exams conducted by the Directorate General of Civil
Aviation (DGCA).
Pass a practical flying test with a DGCA examiner.

You should be academically excellent and have good grades.
Budget Considerations
This is only a rough estimate I could find...

Flying school fees: Between ₹20 lakhs to ₹1cr for the PPL course,
depending on the school and how many flying hours you complete.
Medical exams, ground school, and exam fees: Around ₹1
lakh to ₹5 lakhs
Miscellaneous fees: upto ₹5 lakhs

Most flight schools aren't in urban areas so you need to pay for accommodation too, which is an additional amount here.
In India recreational flying is not too common and almost all pilots get a PPL only to advance to the CPL and make this their career.
Well-known flight schools
There's many flight schools in India that cater to many budget options and offer great services.
You can find them according to your convenience.

Although this answer is for India only, this doesn't exactly vary to much. You can get even more exact details by calling a flight school.


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How can I hear the ATC communications with my instructor loud and clear in my headset
I am training with the DA40 TDI GARMIN 1000 FADEC aircraft, whenever me and my instructor talk to each other,it is loud and clear but when he speaks with ATC or ATC with him, it becomes quiet and unclear. I think it’s an issue with the intercom isolation but I’m not sure. Please can anyone help?

What radio and intercom? Sounds like the sidetone is turned way down.

Have you talked to your instructor about this? He should know how to configure the audio panel so you can hear. Learning to use the radio is part of your training. If you can't hear ATC you can't learn. Also, why is he doing all the radio calls? When I was learning, my instructor was coaching me while I was making most of the calls. He only stepped in if it was something unusual that he wanted.

Yes I did he said it’s just the way the aircraft is built, whoever uses the push to talk button has priority on hearing ATC ♂

You could ask both ATC and your instructor to scream into their respective microphones instead

Generally you have separate volume controls for the intercom (you and you instructor) and the radios (between you and ATC). You should be able to adjust until they are to your liking. Separately, you will develop a skill that mentally mutes your instructor as soon as you hear the first "crackle" of a radio communication. This just takes time, but it will come. The more practice and air-time you get, the less busy and frantic things feel. This allows you time to sort out the finer points beyond just keeping the shiny side up. Good luck.

Check the volume on your radios. If those are low, then it’s not going to be loud and clear in your headset.  Your intercom volume is controlled By the lower knob on the G1000 Audio panel. Your COM radio volumes are controlled by the small knob on the upper right hand corner of both the PFD and MFD displays.


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Trimming direction on climb out and descent to maintain an airspeed
What direction do I trim in order to catch an airspeed with the control stick while on the approach in the circuit pattern? When I normally trim out the control pressures while in the region of normal command it’s simple. see I am on the upwind of the departure leg if I push the nose down and I dream nose down by pushing the trim wheel forward I relieve the push down pressure from the control stick
But if I’m on the descent now it gets a little tricky especially in the traffic pattern down wind, base and approach,were we go into slow flight and region of reverse command I initially push the control stick to an attitude that gives me a higher speed, now what do I do to catch that speed when I get that speed? Do I trim nose down or do I trim nose up after getting that speed?

Thanks guys but do you guys think it’s wrong to do this  I pulled the trim wheel backward for a nose up attitude after pushing the control stick forward to just pass 80 knots in order to get 75 knots for my short final and just waited to see if the attitude and speed settles and it worked, I just now player with the throttle to maintain height as the runway got closer. It got the job done but is this wrong practice?

One- are you really in the region of reverse command at any point before you begin your landing flare?  May vary from one a/c to another.  Also w/ pilot technique, and w/ particular situation (i.e. short/ soft field landing, vs normal, vs gusty crosswind...)
Two-- your a/c's tendency to pitch to a higher or lower a-o-a (i.e. a lower or higher airspeed) as you change power will likely be the dominant issue as far as trim is concerned.
Three-- once you obtain any given desired airspeed, and capture that speed via stick/yoke inputs, the way you maintain that airspeed w/o stick/yoke pressure is simply to trim the stick/yoke pressure to zero.  If pushing, trim nose-down till the stick/yoke pressure goes to zero, and if pulling, trim nose-up till the stick-yoke pressure goes to zero.
Four-- keep in mind that for a normal, stable, aircraft, the "region of reverse command" just means that your sink rate will increase if you slow down without adding power.  It doesn't mean that you'll need less back pressure, or possibly even some forward pressure on the stick/yoke to fly slower without changing the throttle position.  For a normal, stable, aircraft, even in the region of reversed command, starting from any trimmed airspeed, to slow down you need to move the stick/yoke aft by holding aft pressure on the stick/yoke, and to speed up you need to move the stick/yoke forward by holding forward pressure on the stick/yoke.  So apply that pressure on the stick/yoke as needed, and then trim it away.  There won't be anything fundamentally different in the region of reversed command, than in the region of normal command.

Elevator trim sets the hands free angle of attack the airplane will seek on its own, for a given weight, center of gravity, and power configuration. In the absence of an angle-of-attack indicator, we use airspeed as a proxy for angle of attack, since they are directly related.
You generally let the trim do most of the work of holding a desirable pitch attitude for you, except for short term maneuvering inputs. So you should be in the habit of trimming out control pressures to let the airplane do the work of holding a speed/attitude for any case where you need the airplane to fly like that for more than a short time, say, 15 to 30 seconds. That way, if your attention is diverted and you relax your vigilance momentarily, no worries, as the plane will hold the speed/attitude you want largely on its own except for minor deviations due to turbulence or some such. You should think of the trim wheel as a hands-free speed control dial.
So if you're at, say, 70 kt and you are holding forward stick pressure to keep the nose from rising and the airplane slowing down, you trim nose down until the nose stays where you want it hands off. This leaves you free to pay attention to other things while you fly. Don't over think it. If you are holding forward pressure, trim ND, if back pressure trim NU until the nose attitude and speed stay where you want them without your input.
With tractor engine power planes with the engine in the nose, there is an added complication you must account for. Changes in power affect trim, either because the thrust line is above or below the center of mass and causes pitching moments, or because the increase in propeller wash over the tail increases the tail's effectiveness. Generally the change in tail effectiveness dominates, and you find that an increase in power changes trim in the nose-up direction (because the prop is "blowing" on the tail harder).
So, trim speed will often drop when you add power. Experiment yourself in the airplane to see the effect: Set the plane up in level flight, trimmed hands off to hold 80 kt, then add power to climb, and leave the stick alone and just watch what the airplane does. It will hunt up and down for a while, seeking, then settling down at it's original trim state, but now in a climb because the power was increased. But you'll notice the trimmed speed is no longer 80kt. It will be some knots lower, say, 75 kt. Reverse the effect for power reductions. A reduction in power for a tractor engine will result in higher trimmed speed when everything settles down. It's as if a passenger climbed from the front seat to the back seat when you add power, and vise versa.
As you get on top of this phenomenon, you will anticipate it and learn to add a slight amount of nose down trim when adding power to climb (assuming you want to maintain the same speed) in order to keep the airplane from seeking a slower trim speed than you originally set. This is important safety wise in situations like go-arounds where you are firewalling the throttle with the plane trimmed to its final approach speed. If you do nothing, and relaxed on the controls due to some other distraction, the power increase will reduce the trim speed and the pitch up will be more than that required to convert the extra power to climb. You might find yourself approaching stall in a high power climb situation if you weren't paying attention to pitch.

You always pitch down to gain airspeed and pitch up to slow down.  Remember, gravity is usually around 4x the strength of your plane at full throttle!
Trimming for proper airspeed in the pattern is a great way to simplify landing.  Instead of "high, low, fast, or slow?" you now only have "high or low", which can be managed with power.
Use the trim wheel to set your approach speed.  It can be the first thing to do going from downwind to base.  Then look.  You're either high or low.

"gravity is usually around 4x the strength of your plane at full throttle!" I'm probably dumb but what do you mean strength of your plane?? Like the lift? or what?

@AadirajAnil that would be thrust or forward propulsive force.  Gravity contributes to the sine of glide angle (below the horizon) × aircraft weight

"You always pitch down to gain airspeed and pitch up to slow down."  Nope.  Not if you plan to maintain a constant altitude.  To perform a level speed change you lead with power and trim as speed changes.

@MichaelHall the question is $obviously$ about trimming.  Your point is correct, what is the downvote for?  do I trim nose up or down to get that speed? (final line of question).

The DV is for the very reason I mentioned.  You don't "always" pitch to change airspeed.  (absolute statements can be gotchas, especially so when you make them bold!)

But since the question specifically asks about a climb or descent I will cut you some slack -  Get rid of the emphatic "always", or clarify controlling speed with pitch in a climb or descent and I will delete all comments and remove my DV...

@MichaelHall appreciate the comments.  If an aircraft is staticly stable pitch, pitch controls speed, power controls climb, level, or descent.  As I'm sure you know, pitch and power go hand and hand in controlling the aircraft.  (When I was learning aily and rud-rud were friends too).  We all have are ways, and hopefully we all land safe.

Again, nope.  You are thinking in simplistic absolutes.

But, for more advanced training, on less staticly stable aircraft, I totally agree.  A practical application for what you are talking about would be slowing from a pattern speed of 100 knots to approach speed of 65 knots in a 172.  Yes, absolutely, cut power and trim up (dropping flaps 10 will also raise the nose).

But "simplistic absolutes", if they work, need not be changed.  It does (however) depend on what one is flying.

No, it doesn't depend on what you are flying, or on stability either.

Guys, I think in the speed unstable region, which is during slow flight most often in an approach, speed is ultimately controlled by the pitch, and then your power or your altitude is mainly controlled by your power settings, but in the speed stable region, which is often when the aircraft has enough power and enough speed; a scenario is when we cruise, power controls speed and pitch controls altitude that’s why when you’re cruising if you point or pitch your airplane upwards to a certain nose up attitude you climb and  if you pull the power back your speed is gonna reduce in FAST FLIGHT

@Philoko you seem to be on the right track.  It's really a combination of pitch and power.  You can actually do both at the same time (yoke and throttle), for example, to capture a cruising altitude after climbing, THEN trim after you reach cruising speed.

Geez Robert, that's what i have been trying to tell you for at least a couple of years now... (eye roll)

In level flight portions of traffic pattern you set power, then trim to desired airspeed and if necessary, adjust power a bit to maintain speed and altitude.
In descent you again set power, trim for desired airspeed, and then control the rate of decent with power.
As Robert described in his answer, always trim nose up to reduce airspeed and nose down to increase airspeed. You must use the throttle simultaneously to keep the assigned altitude or proper rate of decent.
Stick/yoke should always be trimmed to feel neutral, in traffic pattern you are essentially trimming most of the time


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Hoping to get some details on the GE CF6-50 Turbine bucket thermal protective coating
I just bought this beautiful little bucket and am trying to find the details so I can describe it accurately.

Just based on the gold appearance I assumed it was TiN, but looking further it seems like more exotic coatings are more typical, although I couldn't find any clear references.
Details on the specific Nickel superalloy or other notes (holes are laser or EDM?) would be appreciated.

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.

Hi there, welcome to Aviation.SE! Please read [tour], [help/on-topic], and [help/dont-ask] before posting a question. Although your title state that this is from the General Electric CF6 turbine, and thus related to aviation, not all people here will be able to determine the material of the coating. Please check out Chemistry.SE. The people there may be more suited to answer your question.

@AircraftEnthusiast007 I seriously doubt that Chemistry.SE could help here, just based on a picture. The people who know about materials used in turbine engines are more likely to be found here. And asking questions about which materials are used in aircraft engines is certainly on-topic here.


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For a fixed angle of attack, can increasing your speed cause a stall?
Most sources (I've found while googling) discuss angle of attack as the primary factor for determining if a wing stalls, and that makes sense.  As a layperson I also hear about minimum stall speeds of aircraft but not maximum.  But I've been reading about how air flow detaches from the back of a car at higher speeds so it got me wondering why doesn't that happen with wings? or does it?
In particular, for a fixed angle of attack, why wouldn't an increasing airspeed eventually cause flow to separate and therefore stall the wing?
I realize that near and above the speed of sound weird things happen, but I'm more interested at lower speeds.

Posting this as a comment because it doesn’t really answer your question, but yes, they are called high-speed stalls. For a given angle of attack, the flow could be attached at a lower speed, and then separate at an increased speed.

Thanks.  I've found this question really hard to google but i find 90% is knowing the right words :)

This has to do with the so-called coffin-corner. See this answer, also from @Peter Kämpf. And this also relates to why fast aircraft fly using the Mach number in cruise: What is the reason for changing the speed reference (IAS or Mach number) with altitude?.

Normally, increasing speed reduces the impact of viscosity (expressed as the Reynolds number). Viscosity is the reason why airflow detaches, and less of it means that flow will stay attached to slightly higher angles of attack.
However, transsonic effects spoil this pretty picture: Once supersonic areas form around the wing, raising the speed will increase their size and the shock in which they end. Since a shock is a sudden deceleration of air, and deceleration helps to let flow separation occur, an initially weak shock without separation can become a stronger shock with flow separation downward of the shock when speed is increased. If that separation causes a sufficient reduction in lift or severe buffeting so flight cannot be maintained, a so-called high-speed stall will follow.
Below the transsonic speed range, there is only one edge case where flying faster at the same AoA could increase the risk of stall. But that needs the right definition of speed as true air speed: When climbing, the airplane needs to compensate for the decreasing air density by flying faster. Higher altitude also means lower air temperature, and this effect will reduce the Reynolds number of the flow over the wing, even at the same indicated airspeed. Not by much, but if you start just a bit above stall, the lower Reynolds number at a much higher altitude could help to get you into a stall.
See here for an answer about the change in Re with altitude.

@MichaelHall I've read the "eventually" part as the upper limit of the subsonic speed range. Flying higher gs is indeed the way to keep the same AoA, but then the limit will be structural, not aerodynamic. Higher speed equals higher Reynolds number, except you vary height to keep AoA constant at increasing TAS.

@MichaelHall Yes, the last sentence makes clear that an answer without transsonic effects is preferred. Now I added a paragraph, but to stall an airplane on purpose this way needs very precise flying and still air.

I’m having a hard time understanding why the changing Re number will allow the wing a higher AoA without stalling. The actual viscosity of the air stays the same regardless of speed, correct?

(it could be as simple as more turbulence (higher Re) causes flow to separate later but I’m thinking it’s more complicated)